Show that \(\tilde{H}_n(X) \approx \tilde{H}_{n+1}(SX)\) for all \(n\), where \(SX\) is the suspension of \(X\). More generally, thinking of \(SX\) as the union of two cones \(CX\) with their bases identified, compute the reduced homology groups of the union of any finite number of cones \(CX\) with their bases identified.
Solution
Write the suspension as \[ SX=C_+X\cup C_-X, \qquad C_+X\cap C_-X=X, \] where both cones are contractible.
1. For every \(n\),\[ \boxed{\widetilde H_{n+1}(SX)\cong\widetilde H_n(X).} \]
The reduced Mayer–Vietoris sequence contains \[ \widetilde H_{n+1}(C_+X)\oplus\widetilde H_{n+1}(C_-X) \to\widetilde H_{n+1}(SX) \to\widetilde H_n(X) \to\widetilde H_n(C_+X)\oplus\widetilde H_n(C_-X). \] Both outer groups vanish because cones are contractible, so the middle map is an isomorphism.
Now let \(Y_r\) be the union of \(r\ge1\) cones on \(X\), all with their bases identified.
2. For \(r\ge2\),\[ Y_r\simeq\bigvee^{r-1}SX. \]
Choose one cone \(C_1X\subset Y_r\). It is a contractible subcomplex, so collapsing it to a point is a homotopy equivalence: \[ Y_r\simeq Y_r/C_1X. \] For each remaining cone \(C_iX\), its common base \(X\) has been collapsed to that point. The quotient \[ C_iX/X \] is the suspension \(SX\). Different cones meet only in the collapsed base, so \[ Y_r/C_1X\cong\bigvee_{i=2}^r(C_iX/X) \cong\bigvee^{r-1}SX. \]
3. Hence for every \(k\),\[ \boxed{ \widetilde H_k(Y_r) \cong \bigoplus^{r-1}\widetilde H_{k-1}(X). } \]
Reduced homology sends a finite wedge to the direct sum of reduced homology groups. Apply 2 and then 1 to each suspension summand.